Chapter 8.3, Problem 22E

To determine

To find : the volume of the solid.

Answer to Problem 22E

The volume of the solid is $\pi \left(3\pi -8\right)$ .

Explanation of Solution

Given information : The region in the first quadrant bounded above by the line $y=2$ below by the curve $y=2\sin x,0\le x\le \raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ , and on the left by the y-axis, about the line $y=2$ .

Calculation :

Volume of the solid:

  $\begin{array}{l}V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}{\left(2-2\sin x\right)}^{2}dx}\\ V=\pi {\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(4-8\sin x+{\sin }^{2}x\right)dx}\\ V=4\pi \left[{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(1-2\sin x\right)dx+{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(\sin x\right)dx}}\right]\\ V=4\pi \left[{\left(x+2\cos x\right)}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+{\displaystyle \underset{0}{\overset{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\int }}\left(\frac{1}{2}-\frac{1}{2}\cos 2x\right)dx}\right]\\ V=4\pi {\left[x+2\cos x+\frac{1}{2}x-\frac{1}{4}\sin 2x\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ V=4\pi {\left[\frac{3}{2}x+2\cos x-\frac{1}{4}\sin 2x\right]}_0^{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ V=4\pi \left(\frac{3\pi }{4}-2\right)\\ V=\pi \left(3\pi -8\right)\end{array}$